[Leetcode] 4. Median of Two Sorted Arrays (JAVA)
1755 ワード
Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.
The overall run time complexity should be O(log (m+n)).
Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.
Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3)/2 = 2.5.
Input: nums1 = [0,0], nums2 = [0,0]
Output: 0.00000
Input: nums1 = [], nums2 = [1]
Output: 1.00000
Input: nums1 = [2], nums2 = []
Output: 2.00000
nums1.length == m
nums2.length == n
0 <= m <= 1000
0 <= n <= 1000
1 <= m + n <= 2000
-106 <= nums1[i], nums2[i] <= 106
The overall run time complexity should be O(log (m+n)).
Example 1:
Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.
Example 2:
Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3)/2 = 2.5.
Example 3:
Input: nums1 = [0,0], nums2 = [0,0]
Output: 0.00000
Example 4:
Input: nums1 = [], nums2 = [1]
Output: 1.00000
Example 5:
Input: nums1 = [2], nums2 = []
Output: 2.00000
Constraints:
nums1.length == m
nums2.length == n
0 <= m <= 1000
0 <= n <= 1000
1 <= m + n <= 2000
-106 <= nums1[i], nums2[i] <= 106
My Code
class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int len1=nums1.length, len2=nums2.length;
int i1=0, i2=0;
int[] arr = new int[(len1+len2)/2+1];
for(int i=0 ; i<arr.length ; i++) {
if(i1<nums1.length && i2<nums2.length) {
if(nums1[i1]<=nums2[i2]) {
arr[i] = nums1[i1++];
} else {
arr[i] = nums2[i2++];
}
} else {
if(i1<nums1.length) {
arr[i] = nums1[i1++];
} else {
arr[i] = nums2[i2++];
}
}
}
return (arr[(len1+len2-1)/2] + arr[(len1+len2)/2])/2.0;
}
}
Reference
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